In this vague formulation (without any example!), it's hard to be helpful. The only related optimization problem is Lagrange multipliers (en.wikipedia.org/wiki/Lagrange_multipliers) and its variational generalization, Pontryagin's principle.
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Wadim ZudilinJun 29 '10 at 9:32

@Wadim - Yes it is vague. This is very much outside my area of expertise and I don't really know the correct terminology to express the problem, for that I apologize. @muad - Thank you for the suggestion. Glancing at the formulation of dynamically billiards it looks like a promising route, although admittedly I am unsure of how to translate a problem expressed in terms of energy to a problem where the conditions are expressed in velocity and position.
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Jonathan FischoffJul 5 '10 at 8:15

A bit confused: the box is one dimensional? (Since you write l < y(t) < u) Then can't you use the fundamental theorem and conclude that for any path, J(y) = c - d? Or do you actually want to minimize |y''| instead?
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Willie WongJul 5 '10 at 10:59

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@muad: billiard is probably not the right potential to use, at least by itself. If the solution bounces off the walls then the second derivative of y cannot be square integrable (as it'd contain a delta function). So you probably need condition that whenever $y(t) = l \implies y'(t) = 0, y''(t) \geq 0$ and similarly for the upper bound.
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Willie WongJul 5 '10 at 18:32

1 Answer
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Just to elaborate a bit on what Rahul and I mentioned in the comments.

Take the action functional to be $\int_0^1 (y'')^2 dt$, with prescribed boundary conditions $y(0) = a$, $y(1) = b$, $y'(0) = c$, $y'(1) = d$. For finding the free evolution, take the variation of the function relative to $y$ and set it to zero. Immediately this gives
$$\int_0^1 y'' (\delta y)'' dt = 0 $$
for any perturbation. The boundary conditions prescribed implies that $\delta y(0) = \delta y(1) = \delta y'(0) = \delta y'(1) = 0$. So we are allowed to integrate by parts twice (assuming the solution is $C^4$) and obtain $y'''' = 0$, which implies that $y(t)$ is a cubic polynomial in time and thus has 4 free parameters, which we can fix by the boundary values.

The intuition for the bounded case is that, until the evolution hits the wall, locally the equation of motion should be identical to the free evolution. So the solution should be composed piecewise of cubic polynomials. Every time it hits the wall it should receive a hard impact, which suggests that $y''''(\tau_i) = c\delta_{\tau_i}$, the Dirac delta. (The same way that for a hard billiard which away from the walls travel via $x'' = 0$ receive a delta function impact in the second derivative when it hits a wall.) This suggests that $y'''$ is a step function of $t$, and $y''$ is continuous.

First solve the free problem: we want
$$ k_3 t^3 + k_2 t^2 + k_1 t + k_0 = y(t) $$
Plugging in the values for the four points we find by solving the linear system that the equation should be
$$ y(t) = 3t^3 - 9 t^2 + 6t $$
which achieves its local max and min in $[0,2]$ at $1\pm \sqrt{1/3}$, at which points $|y| = |\pm \sqrt{4/3}| > 1$. So the free evolution is no go.

A direct computation (if I did it right, which is not guaranteed) with 2 break points $\sigma\in [1,2]$ and $\tau \in[0,1]$ yields that $\tau = 1/2$ and $\sigma = 3/2$ an admissible pair. It is just linear algebra in the end.